Derivation of Time in Air of a Projectile
| Ttop = mgH1 + 1/2 m(Vtop)2 | Energy is Conserved |
| Tbottom = mgH2 + 1/2 m(Vbottom)2 | H2 = 0 (zero level) |
| mgH = 1/2 mVbottom2 | Vtop = 0 (y - direction) |
| 1/2Vbottom2 = gH | Divide by m; m cancels |
| V2 = 2gH | Multiply both sides by 2 |
| V = sqr. Root(2gH) | Square root both sides |
| Vf = Vo + at | Motion equation;Vo = 0; subst. |
| 2gH = (gt)2 | a = gravity;square both sides |
| 2H/g = t2 | divide both sides by g2 |
| sqr. Root(2H/g) = t | square root both sides |
t = sqr. root(2H/g)
| Vf = Vo + at | Motion equation;Vf = 0 - object |
| 0 = Vo + gt | is not moving up/dwn at its apex |
| -Vo = gt | a = gravity;Subtract Vo from |
| -Vo/g = t | both sides and divide by g t = time to get to the top; since its on level ground - tdown = tup |
t = -2Vo / g